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Consider:

$${1 \over 2}\left[ {{e^{3x}} - {e^{2x}} + {e^x} - 1} \right] = \left( {\sum\limits_{k \ge 0} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {1 \over 2}$$

I understood that because of the expression $1\over 2$, One should define $a_k$ as

$${a_k} = \left\{ {\matrix{ {{{{3^k} - {2^k} + 1} \over {2\cdot k!}},k > 0} \cr {0,\qquad \qquad \,\,\,\, k = 0} \cr } } \right.$$

Can you clarify this for me? I understood that $1\over2$ should be treated as a power-series, but how exactly?

AnnieOK
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  • In the definition of $a_k$ don't you need to write also the $k!$ in the denominator? – Jimmy R. Mar 18 '14 at 19:34
  • You right. I'll correct that. – AnnieOK Mar 18 '14 at 19:38
  • There is an idea of writting $1$ as $e^{0x}$. But then you need a definition of $0^0$...Do you see it? – Jimmy R. Mar 18 '14 at 19:44
  • Why should you treat $\frac12$ as a power series? For example, more generally, when you take a power series and you subtract a term or a few (or a polynomial), to get an expression like $$\sum_{k\ge 0}b_k x^k - \sum_{k=0}^{n} c_k x^k,$$ the coefficient of $x^k$ is $b_k$ when $k > n$, and $b_k - c_k$ when $0 \le k \le n$. (What you have is the same, with $b_k = \frac{3^k - 2^k + 1}{2\cdot k!}$, and $n = 0$ and $c_0 = \frac12$.) – ShreevatsaR Mar 19 '14 at 04:23

1 Answers1

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The RHS can be written as $$\begin{align*} \left( {\sum\limits_{k \ge 0} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {1 \over 2}&= \left( {\sum\limits_{k \ge 0} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {1 \over 2}x^0=\\&= \left( {\sum\limits_{k \ge 1} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {3^0-2^0+1-1 \over 2\cdot0!}x^0=\\&=\left( {\sum\limits_{k \ge 1} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - 0\cdot x^0 \end{align*}$$ so that the definition of $a_k$ is now straightforward.

Jimmy R.
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